[mercury-users] New to mercury: structure reuse and efficiency

[Ralph Becket]

Gregory D. Weber, Wednesday, 30 June 2004:
> There's an implied question here which I don't think has been answered:
> how _does_ he do what he wants to do here, which, I believe, is to
> make sure that H does not equal T before making the recursive call?
>
> Isn't the solution to write -- ?
> 
> member([H | T], X) :-
>   (
>     H = X ->
>     true
>   ;
>     member(T, X)
>   ).
> 
> or, equivalently,
> 
> member([H | T], X) :-
>   (
>     if H = X 
>     then true
>     else member(T, X)
>   ).

Yes, either of those will do the trick.

-- Ralph
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